Question: Solve for $q$, $ \dfrac{4q - 3}{q^2} = \dfrac{10}{3q^2} + \dfrac{3}{3q^2} $
Answer: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $q^2$ $3q^2$ and $3q^2$ The common denominator is $3q^2$ To get $3q^2$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ \dfrac{4q - 3}{q^2} \times \dfrac{3}{3} = \dfrac{12q - 9}{3q^2} $ The denominator of the second term is already $3q^2$ , so we don't need to change it. The denominator of the third term is already $3q^2$ , so we don't need to change it. This give us: $ \dfrac{12q - 9}{3q^2} = \dfrac{10}{3q^2} + \dfrac{3}{3q^2} $ If we multiply both sides of the equation by $3q^2$ , we get: $ 12q - 9 = 10 + 3$ $ 12q - 9 = 13$ $ 12q = 22 $ $ q = \dfrac{11}{6}$